A man with mass m1 = 51 kg stands at the left end of a uniform boat with mass m2

Question

A man with mass m1 = 51 kg stands at the left end of a uniform boat with mass m2 = 171 kg and a length L = 3.5 m. Let the origin of our coordinate system be the man’s original location as shown in the drawing. Assume there is no friction or drag between the boat and water.
1)What is the location of the center of mass of the system? 1.35 m
2)If the man now walks to the right edge of the boat, what is the location of the center of mass of the system? 1.35 m
3)After walking to the right edge of the boat, how far has the man moved from his original location? (What is his new location?) 2.7 m
4)After the man walks to the right edge of the boat, what is the new location the center of the boat? 0.946m
5)Now the man walks to the very center of the boat. At what location does the man end up? 1.35m

I have the answers here but I don't understand how you get them!!! What are the formulas and whats the reasoning behind them.

Explanation / Answer

I had solved it before for m1 = 51 kg m2 = 168 kg a length L = 3.1? Just change these to your values. 1) Let the original position of man be labeled as origin, (distance= 0). Center of mass of boat will be at a distance of (3.1)/2 = 1.55. Location of center of mass will be at a distance of say +c m c = (168*1.51)/(168+51)= 1.16 m 2) as there is no external force on the system as a whole, the center of mass has to remain at the same position. As man has reached the other end of the boat. One can draw the diagram of the boat and man in such a way that his new position is symmetrically situated with respect to the center of mass at +1.16 m. Man would move 1.16*2 = 2.32 m to the right and his new location is 2.32 to right of his original position. 3) New location of the center of boat has to be 1.55 m away towards left of his new location. So new location of boat 2.32 - 1.55 = 0.77 m So boat has moved to the left through 1.55 - 0.77 = 0.78 or 0.77 m itself situation has to be symmetric.

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